ACID-BASE EQUILIBRIA (Part One) A Competition for Protons ADEng. PROGRAMME Chemistry for Engineers Prepared by M. J. McNeil, MPhil.

ACID-BASE EQUILIBRIA (Part One) A Competition for Protons ADEng. PROGRAMME Chemistry for Engineers Prepared by M. J. McNeil, MPhil. Department of Pure and Applied Sciences Portmore Community College Main Campus 1

LECTURE OBJECTIVES (PART A) 2

LECTURE OBJECTIVES (PART B) 3

REVIEW How would you define an acid? How would you recognize an acid? How would you define a base? How would you recognize a base? 4

THREE DEFINITIONS OF ACIDS Who Theory: Acid= When Arrhenius increases H + 1880 s Brønsted proton donor (for bases?) 1923 Lowry ditto 1923 Lewis Electron-pair acceptor 1923 5

HYDRONIUM SPECIES Brønsted-Lowry definitions state that acid-base reactions involve the from one substance to another. A hydronium ion, H 3 O +, (a hydrated H + ) is simply a proton with no surrounding valence electrons. In practice, H + ions do not exist in solution and are more likely to exist as water H 3 O + Remember - acids donate and bases accept! Hydronium ion When water accepts a proton from an acid, the product is a hydronium ion. (H 3 O + ) Hydronium ions are represented by either the H 3 O + (aq) or H + (aq) 6

BRØNSTED-LOWRY THEORY OF ACIDS-BASES DEFINITION Acid - Proton Donor An acid is a proton donor, any species that donates an H + ion. Acids donate a proton to water An acid must contain H in its formula; HCl, HNO 3 and H 2 PO 4 - are two. A Brønsted Lowry acid must have a removable (acidic) proton. HCl, H 2 O, H 2 SO 4 A Brønsted Lowry base must have a pair of nonbonding electrons. NH 3, H 2 O Base - Proton Acceptor A base is a proton acceptor, any species that accepts an H + ion. A base must contain a lone pair of electrons to bind the H + ion; Bases accept a proton from water A few examples are NH 3, CO 3 2-, F -, as well as OH -. Therefore, in the Brønsted-Lowry perspective, one species donates a proton and another species accepts 7 it: an acid-base reaction is a proton transfer process.

HYROGEN CHLORIDE AS A BRØNSTED-LOWRY ACID Brønsted-Lowry: must have both an Acid: Proton donor and a Base: Proton acceptor According to Brønsted-Lowry, when hydrogen chloride reacts with water, a proton is transferred from the hydrogen chloride molecule to a water molecule, forming a hydronium ion and a chloride ion. Hydrogen chloride acts as Brønsted-Lowry acid; water acts as Brønsted-Lowry base. 8

AMMONIA AS A BRØNSTED-LOWRY BASE According to Brønsted-Lowry, when ammonia reacts with water, a water molecule acts as a Brønsted-Lowry acid donating a proton to ammonia, the Brønsted-Lowry base. Rule of thumb: Protons may be gained in a reaction with one substance, but lost in a reaction with another substance. What is the name of substances that can operate as both Brownsted Lowry acid and base? HCO 3-9 Brønsted-Lowry acids and bases are always paired.

ACID-BASE EQUILIBRIUM A Competition for Protons When NH 3 (g), a basic substance reacts with HCl(g), an acidic substance, a neutralization occurs (in the gas state) that does not involve hydronium ion or water. In this reaction, an H + ion transfers from Cl atom in the HCl molecule to the N atom of the NH 3 molecule. According to Brønsted-Lowry concept, acid-base reactions involve the transfer of a proton. These reactions are universally reversible and result in an acidbase equilibrium. 10

CONJUGATE ACIDS AND BASES From the Latin word conjugare, meaning to join together. Reactions between acids and bases always yield their conjugate bases and conjugate acids. There will always be two acids (name them) and two bases (name them) in a typical Bronsted-Lowry acid-base reaction. Consider the reverse reaction. How are they all formed? Conjugate acid-base pair two substances that differ only by one H + unit. In an acetic acid solution, we can describe the forward reaction as proton transfer from acetic acid to water molecules. What about the reverse? The acid is the more positive species having an extra 11 H.

WHAT HAPPENS WHEN AN ACID DISSOLVES IN WATER? Water acts as a Brønsted- Lowry base and abstracts a proton (H + ) from the acid. As a result, the conjugate base of the acid and a hydronium ion are formed. 12

CONJUGATE ACIDS AND BASES 13

REVERSIBLE ACID-BASE REACTIONS Which is the acid and which is the base in each of these rxns? Which is the conjugate acid and which is the conjugate base in each of these rxns? Consider the reverse reaction. 14

IDENTIFYING CONJUGATE ACID-BASE PAIRS The following chemical reactions are important for industrial: (a) HSO 4 - (aq) + CN - (aq) SO 4 2- (aq) + HCN (aq) (b) ClO - (aq) + H 2 O (l) HClO (aq) + OH - (aq) (c) S 2- (aq) + H 2 O (aq) HS - (aq) + OH - (aq) Plan: To find the conjugate acid-base pairs, we find the species that donate H + and those that accept it. The acid (or base) on the left becomes its conjugate base (or acid) on the right. (Try for homework aminoethane dissolved in water.) 15

SOLUTIONS (a) The proton is transferred from the sulfate to the cyanide so: HSO 4 - (aq) /SO 4 2- (aq) and CN - (aq)/hcn (aq ) are the two acid-base pairs. (b) The water gives up one proton to the hypochlorite anion so: ClO - (aq)/hclo (aq) and H 2 O (l) / OH - (aq ) are the two acid-base pairs. (c) One of water s protons is transferred to the sulfide ion so: S 2- (aq)/hs - (aq) and H 2 O (l) /OH - (aq) are the two acid-base pairs. 16

ELECTROLYTES - A REVIEW What are electrolytes? Electrolytes can be classified based on their strengths. 1. What are strong electrolytes? 2. What are weak electrolytes? A solution can conduct electricity owing to the quantity of ions it contain. Relationship between strong/weak acids/bases and strong/weak electrolytes. Consider pure water (few ions). 3. What are non-electrolytes? 4. Give two examples in each case. 17

THE AUTO-IONIZATION OF WATER In pure water, a few molecules act as bases and a few act as acids. The two molecules are said to react to form H 3 O + and OH -. The reaction of an acid HA with water to form H 3 O+ and a conjugate base. Water is amphoteric. Acid Base Conjugate Conjugate 18 acid base

ION-PRODUCT CONSTANT FOR H 2 O (K w ) Because the auto-ionization of water is an equilibrium process, there is an equilibrium-constant expression: H 2 O (l) + H 2 O (l) H 3 O + + OH - The ion-product for water, K w : K c = [H 3 O + ][OH - ] [H 2 O] 2 K w = [H 3 O + ][OH - ] = 1.0 x 10-14 can also be written as [H + ][OH - ] = 1.0 x 10-14 For pure water the concentration of hydroxyl and hydronium ions must be equal: [H 3 O + ] = [OH - ] = 1.0 x 10-14 M = 1.0 x 10-7 M (at 25 C) K c [H 2 O] 2 (are both constants) = K w K w = [H 3 O + ][OH - ] = 1.0 x 10-14 (at 25 C) 19

RELATIVE STRENGTHS OF ACIDS AND BASES STRONG ACIDS A strong acid completely transfers its protons to water, leaving no undissociated molecules in water. It totally dissociates in water. 7 most common strong acids include 6 monoprotic acids (HCl, HBr, HI, HNO 3, HClO 3, and HClO 4 ) and one diprotic acid, H 2 SO 4. WEAK ACIDS A weak acid only partially dissociates in water, and exists as a mixture of acid molecules and their constituent ions. The conjugate base of a weak acid is a weak base. In every acid-base reaction, the position of equilibrium favors transfer of the proton from the stronger acid to the stronger base to form the weaker acid and the weaker base. 20

RELATIVE STRENGTHS OF ACIDS The behavior of acids of different strengths in aqueous solution. The extent of dissociation is shown below. 21

RELATIVE STRENGTHS OF BASES STRONG BASES A strong base is therefore WEAK BASES A weak base is therefore Strong bases are the soluble hydroxides, which are the alkali metal (NaOH, KOH)and heavier alkaline earth metal hydroxides (Ca(OH) 2, Sr(OH) 2, and Ba(OH) 2 ). Weak bases. Many compounds with an electron-rich nitrogen are weak bases (none are Arrhenius bases). The common structural feature is an N atom that has a lone electron pair in its Lewis structure. 1) Ammonia (:NH 3 ) 2) Amines (general formula RNH 2, R 2 NH, R 3 N), such as CH 3 CH 2 NH 2, (CH 3 ) 2 NH, (C 3 H 7 ) 3 N, and C 5 H 5 N 22

ACID-BASE CHARACTER AND THE ph SCALE 23

MEASURING THE ph OF AN AQUEOUS SOLUTION (1) ph paper (less precise) (2) Electrodes of a ph meter 24

OTHER METHODS OF MEASURING ph (3) Indicators can also be used, but they are less precise. 25

ph SCALE ph = -log[h + ] poh = -log[oh + ] Neutral solution: ph = -log(1.0 x 10-7 ) = -(-7.00) = 7.00 The ph decreases as the [H + ] increases. 26

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CALCULATING ph, [H 3 O + ], [OH - ] and poh In acidic solutions, the protons that are released into solution will not remain alone due to their large positive charge density and small size. They are attracted to the negatively charged electrons on the oxygen atoms in water, and form hydronium ions. H + (aq) + H 2 O (l) = H 3 O + (l) [H + ] = [H 3 O + ] To handle the very large variations in the concentrations of the hydrogen ion (cumbersome) in aqueous solutions, the ph scale is used which is: ph = - log[h 3 O + ] (Definition) N.B. In general, a lowercase p before the symbol means negative of the symbol. ph is the negative logarithm of the molar concentration of H + or H 3 O + ions. Analyzing problems regarding chemical equilibrium solutions of acids and bases, you must account for all the entities that may contribute to the solution s H + and OH - ions concentrations since these give rise to the acidbase properties of the solution. 28

CALCULATING ph OF A STRONG ACID Question 1. Calculate the ph of 0.001 M sulphuric acid. H 2 SO 4 (aq) 2H + (aq) + SO 2-4 (aq) Initial: 0.001 M 0 M 0 M Change: -0.001 M +2(0.001 M) +0.001 M Equlibirum: 0 M 0.002 M 0.001 M ph = - log[h + ] (scale) ph = -log [0.002 M] ph = 2.69 Because strong acids completely ionize (100%), the [H + ] of a solution made with a strong acid is easily figured out, since it is equal to the molarity of the solution. Thus, a 0.1 M solution of HCl has a [H + ] of 0.1 M. A 0.029 M has a [H + ] is 0.029 M etc. Question 2. Calculate the [H + ] of a solution that has a ph of 4.3. (Ans. 5.0 x 10-5 M) 29

CALCULATING ph OF A STRONG ACID Question 3. What is the ph of a solution that is 10-12 M in hydronium ion? ph = -log[h 3 O + ] = ph = (-1)log 10-12 ph = (-1)(-12) ph = 12 ph of a neutral solution = 7.00 ph of an acidic solution < 7.00 ph of a basic solution > 7.00 Question 4. What is the ph of a solution that is 7.3 x 10-9 M in H 3 O +? ph = -log(7.3 x 10-9 ) ph = -1(log 7.3 + log 10-9 ) ph = -1[(0.863)+(-9)] ph = 8.14???? Question 5. What is the ph of a 0.040 M solution of HClO 4? ph = -log(0.040) ph = 1.40 Why is the ph of 1 x 10-8 M HNO 3 is 6.963 and not 8.00? Question 6. What is the of a 2.0 x 10-8 M HCl solution? 30

CALCULATE [H 3 O + ], ph, [OH - ] and poh OF A STRONG ACID Question 7: A chemist dilutes concentrated hydrochloric acid to make two solutions: (a) 3.0 M and (b) 0.0024 M. (c) Calculate the [H 3 O + ], ph, [OH - ], and poh of the two solutions at 25 C. Plan: Solution: (a) [H 3 O + ] = 3.0 M We know that hydrochloric acid is a strong acid, so it dissociates completely in water; therefore [H 3 O + ] = [HCl] initial. We use the [H 3 O + ] to calculate the [OH - ] and ph as well as poh. ph = -log[h 3 O + ] = -log(3.0) = K [OH - ] = w 1 x 10 = -14 = M [H 3 O + ] 3.0 poh = - log(3.333 x 10-15 ) = 15.000-0.477 = K w = [H 3 O + ] [OH - ] for calculation of strong acids (b) [H 3 O + ] = 0.0024 M ph = -log[h 3 O + ] = -log(0.0024) = K [OH - ] = w 1 x 10 = -14 = M [H 3 O + ] 0.0024 poh = -log(4.167 x 10-12 ) = 12.000-0.6198 = 31

CALCULATING [H 3 O + ], ph, [OH - ] and poh OF A STRONG BASE Question 8. Calculate the ph of a 0.030 M barium hydroxide. Ba(OH) 2 (aq) Ba 2+ (aq) + 2OH - (aq) Initial: 0.030 M 0 M 0 M Change: -0.030 M +0.030 M +2(0.030 M) Equlibirum: 0 M 0.030 M 0.060 M poh = - log[oh - ] (scale) poh = -log [0.060 M] ph = 1.0 But, ph + poh = 14 ph = 14-1 ph = 13.0 Alternative K w = [H 3 O + ] [OH - ] for calculation of strong bases too. Question 9. Calculate the H 3 O +,ph, [OH - ] and poh for a 0.015 M calcium hydroxide solution. [OH-] = 0.030 M ph = 12.48 H 3 O + = 3.3 x 10-13 M 32

CALCULATING THE ph OF WEAK ACIDS AND WEAK BASES WHEN ASKED TO CALCULATE THE ph OF A WEAK ACID OR A WEAK BASE, YOU MUST BE GIVEN THE K a or K b OF THE ACID AND BASE, RESPECTIVELY. 33

ACID AND BASE DISSOCIATION CONSTANTS The percentage of the molecules that break apart (ionizes) depends on the weak acid. Some weak acids dissociate very minimally, and as a consequence such an acid has little acid strength. Others dissociate more (more acid strength). The degree of dissociation of a weak acid in water is described as the acid dissociation constant, K a. (Comparing the relative strengths of weak acids.) K a is the equilibrium constant for the ionization of a weak acid in water. The base ionization constant (K b ) is the equilibrium constant for a weak base. The numerical values tell to what extent ions are formed from the molecules of the acid or base. 34

CALCULATING THE ph OF WEAK ACIDS AND WEAK BASES WEAK ACIDS DISSOCIATION CONSTANT WEAK BASES DISSOCIATION CONSTANT 35

ACID-IONIZATION EQUILIBRIA The reaction of a weak acid (or base) with water is the same as previously discussed. [ H3O ][ A HA( aq) H 2O( l) H3O ( aq) A ( aq) Ka [ HA] Hydronium ion concentration must be determined from the equilibrium expression. ] Relative strengths of weak acids can be determined from the value of the equilibrium constant. Small values of K a mean that the water molecules solution has relatively large number of molecules and few ions. Large equilibrium constant, K a, means strong acid (more acid dissociates to form H + ) Small equilibrium constant means weak acid (less acid dissociates to form H + E.g. Determine which acid is the strongest and which the weakest. Give reasons. Acid K a HCOOH 1.8x10 4 CH 3 COOH 1.8x10 5 HF 3.5x10 4 NEVER USE ph AS A COMPARISON OF ACID STRENGTH. ALWAYS AND FOREVER USE K a. Chapter 17-36

K a FOR COMMON MONOPROTIC ACIDS The greater the value of K a, the stronger the acid. [H 3 O + ][A - ] [HA] 37

K b FOR COMMON BASES K b can be used to find [OH ] and, through it, ph. Weak bases include NH 3, amines and the anions of weak acids. 38

pk a SCALE K a values for weak acids tend to be very small number (exponents) and cumbersome to work with. K a values are usually converted to pk a values similar to how [H + ] values are converted to ph (scale). Same goes for K b. NEVER USE ph AS A COMPARISON OF ACID STRENGTH. ALWAYS AND FOREVER USE K a. pk a = - log K a (Definition) Example: Calculate the pk a for acetic acid given that the K a is 1.74 x 10-5 M. pk a = -log [1.74 x 10-5 M] pk a = 4.76 The acid with the smaller is pk a ALWAYS the stronger acid. The converse is true. What s the relationship between K a and pk a? The sample principle applies to pk b. Question 9. Oxalic acid has K a = 6 x 10-2. What is the value of pk a for this acid. 39

CALCULATING ph FROM K a Question 10. Calculate the ph of a 0.30 M solution of acetic acid, C 2 H 3 O 2 H, at 25 C. K a for acetic acid at 25 C is 1.8 10-5 Is acetic acid more or less ionized than formic acid (ants bite) (K a =1.8 x 10-4 )? 40

CALCULATING ph FROM K a Use the ICE table: [C 2 H 3 O 2 ], M [H 3 O + ], M [C 2 H 3 O 2 ], M Initial 0.30 0 0 Change x +x +x Equilibrium 0.30 x x x 41

CALCULATING ph FROM K a Use the ICE table: [C 2 H 3 O 2 ], M [H 3 O + ], M [C 2 H 3 O 2 ], M Initial 0.30 0 0 Change x +x +x Equilibrium 0.30 x x x Simplify: how big is x relative to 0.30? 42

CALCULATING ph FROM K a Use the ICE table: [C 2 H 3 O 2 ], M [H 3 O + ], M [C 2 H 3 O 2 ], M Initial 0.30 0 0 Change x +x +x Equilibrium 0.30 x 0.30 x x Simplify: how big is x relative to 0.30? 43

CALCULATING ph FROM K a Now, x is so so small (<<) that 0.30 x ~ 0.30 M (1.8 10-5 ) (0.30) = x 2 5.4 10-6 = x 2 2.3 10-3 = x ph = log [H 3 O + ] ph = log (2.3 10 3 ) = 2.64 Check: is approximation ok? 44

CALCULATING K a FROM ph Question 11. The ph of a 0.10 M solution of formic acid, HCOOH, at 25 C is 2.38. Calculate K a for formic acid at this temperature. We know that 45

CALCULATING ph FROM K a The ph of a 0.10 M solution of formic acid, HCOOH, at 25 C is 2.38. Calculate K a for formic acid at this temperature. To calculate K a, we need all equilibrium concentrations. We can find [H 3 O + ], which is the same as [HCOO ], from the ph. 46

CALCULATING ph FROM K a ph = -log [H 3 O + ] -2.38 = log [H 3 O + ] 10-2.38 = 10 log [H 3O+] = [H 3 O + ] 4.2 10-3 = [H 3 O + ] = [HCOO ] 47

CALCULATING ph FROM K a In table form: [HCOOH], M [H 3 O + ], M [HCOO ], M Initially 0.10 0 0 Change 4.2 10-3 +4.2 10-3 +4.2 10-3 At Equilibrium 0.10 4.2 10-3 4.2 10-3 4.2 10-3 = 0.0958 = 0.10 48

CALCULATING ph FROM K a K a = [4.2 10-3 ] [4.2 10-3 ] [0.10] = 1.8 10-4 now calculate the ph. 49

CHUCK OFF TIME 1. If K a for NH 4 + is 5.7 x 10-10. Calculate the ph soluton prepared by dissolving 10.69 g off NH 4 Cl in 100 cm 3 of distilled water. (Ans. = 4.47) 2. Nicotinic acid is a weak monoprotic organic acid that we can represent as HA. A dilute solution of nicotinic acid was found to contain the following concentrations at equilibrium at 25 o C. What is the value of K a? [HA] = 0.049 M (Ans. = 4.85) 3. Calculate the concentrations of the various species in 0.10 M hypochlorous acid. K a for hypochlorous acid is 3.5 x 10-8. 4. Consult the past papers. 50

USING K b to CALCUILATE OH - NH 3 (aq) + HOH(l) NH 4+ (aq) + OH - (aq) K b = [NH 4+ ][OH - ] [NH 3 ] = 1.8 x 10-5 Example : Calculate the ph of a 1.00 M Solution of Nitrous acid HNO 2. Solution: HNO 2 (aq) H + (aq) + NO - 2 (aq) K a = 4.0 x 10-4 51

K a AND K b RELATIONSHIP The relationship between the strength of an acid and the strength of its conjugate base is expressed by the equation: K a x K b = K w K a and K b are linked: Combined reaction =? 52

RELATION BETWEEN K a AND K b K a and K b are always inversely related to each other in aqueous solutions. HA(aq)+ H 2 O(l) H 3 O + (aq) + A (aq) Add A (aq)+ H 2 O(l) HA(aq) +OH (aq) 2H 2 O(l) Inverse relationship explains why conjugate base of very weak acid is relatively strong. E.g. given the K a s of the following acid list their conjugate bases in terms of relative strength. K a K b [A ][H 3 [HA ] [HA ][OH [A ] H 3 O + (aq) +OH (aq) K w = K a K b O Acid K a HF 3.5x10 4 HCOOH 1.8x10 4 HOCl 3.5x10 8 HCN 4.9x10 10 ] ] Chapter 17-53

THE RELATION BETWEEN K a and K b OF A CONJUGATE ACID-BASE PAIR Acid HA + H 2 O H 3 O + + A - Base A - + H 2 O HA + OH - 2 H 2 O H 3 O + + OH - [H 3 O + ] [A - ] [H 3 O + ] [OH - ] = x [HA] x [HA] [OH - ] [A - ] K w = K a x K b K a = 4.5 x 10-4 For HNO 2 K a x K b = (4.5 x 10-4 )(2.2 x 10-11 ) = 9.9 x 10-15 K b = 2.2 x 10-11 or ~ 10 x 10-15 = 1 x 10-14 = K w 54

K a and K b K a and K b are related in this way: K a K b = K w Therefore, if you know one of them, you can calculate the other. 55

CHUCKING OFF TIME 1. Niacin, one of the B vitamins, has the following molecular structure: A 0.020 M solution of niacin has a ph of 3.26. (a) What percentage of the acid is ionized in this solution? (b) What is the acid-dissociation constant, K a, for niacin? 2. Calculate the percentage of HF molecules ionized in (a) a 0.10 M HF solution, (b) a 0.010 M HF solution. K a for HF is 6.8 x10-4. 56

DETERMINING ph from K b and INITIAL [B] I Problem 1: Ammonia is commonly used cleaning agent in households and is a weak base, with a K b of 1.8 x 10-5. What is the ph of a 1.5 M NH 3 solution? Plan: Ammonia reacts with water to form [OH - ] and then calculate [H 3 O + ] and the ph. The balanced equation and K b expression are: NH 3 (aq) + H 2 O (l) NH 4 + (aq) + OH - (aq) K b = [NH 4+ ] [OH - ] [NH 3 ] Concentration (M) NH 3 H 2 O NH 4 + OH - Initial 1.5 ---- 0 0 Change -x ---- +x +x Equilibrium 1.5 - x ---- x x making the assumption: since K b is small: 1.5 M - x = 1.5 M 57

DETERMINING ph from K b and INITIAL [B] I Substituting into the K b expression and solving for x: K b = [NH 4+ ] [OH - ] (x)(x) = [NH 3 ] 1.5 = 1.8 x 10-5 x 2 = 2.7 x 10-5 = 27 x 10-6 x = 5.20 x 10-3 = [OH - ] = [NH 4+ ] Calculating ph: [H 3 O + ] = K w 1.0 x 10-14 = [OH - ] 5.20 x 10-3 = 1.92 x 10-12 ph = -log[h 3 O + ] = - log (1.92 x 10-12 ) = 12.000-0.283 ph = 58

Summary - GENERAL STRATEGIES FOR SOLVING ACID-BASE PROBLEMS Think Chemistry - Focus on the solution components and their reactions. It will almost always be possible to choose one reaction that is the most important. Be systematic - Acid-Base problems require a step-by-step approach. Be flexible. Although all acid-base problems are similar in many ways, important differences do occur. Treat each problem as a separate entity. Do not try to force a given problem to match any you have solved before. Look for both the similarities and the differences. Be patient - The complete solution to a complicated problem cannot be seen immediately in all its detail. Pick the problem apart into its workable steps. Be confident - Look within the problem for the solution, and let the problem guide you. Assume that you can think it out. Do not rely on memorizing solutions to problems. In fact, memorizing solutions is usually detrimental, because you tend to try to force a new problem to be the same as one you have seen before. Understand and think; don t just memorize. 59

Summary - GENERAL STRATEGIES FOR SOLVING ACID-BASE PROBLEMS 1. List the major species in the solution. 2. Choose the species that can produce H +, and write balanced equations for the 3. reactions producing H +. 4. Comparing the values of the equilibrium constants for the reactions you have 5. written, decide which reaction will dominate in the production of H +. 6. Write the equilibrium expression for the dominant reaction. 7. List the initial concentrations of the species participating in the dominate 8. reaction. 9. Define the change needed to achieve equilibrium; that is, define x. 10. Write the equilibrium concentrations in terms of x. 11. Substitute the equilibrium concentrations into the equilibrium expression. 12. Solve for x the easy way-that is, by assuming that [HA] 0 x = [HA] 0 13. Verify whether the approximation is valid ( the 5% rule is the test in this case). 14. Calculate [H + ] and ph. 60

INDICATORS and ph CURVES Acid-base indicators are water-soluble weak organic acids (or its conjugate base) or a weak bases (or its conjugate acid) and are used to signal the end of acid-base titrations. Their acid and base forms have different and intense colors, and there is an equilibrium between the two forms: HIn(aq) + H 2 O(l) H 3 O + (aq) + Ind - (aq) acid conjugate base ccid form base form colour 1 color 2 undissociated dissociated 61

Indicators as weak acids Litmus. LITMUS Litmus is a weak acid. It has a seriously complicated molecule which we will simplify to HLit. The "H" is the proton which can be given away to something else. The "Lit" is the rest of the weak acid molecule. HLit (aq) + H 2 O(l) H 3 O + (aq) + Lit - (aq) The un-ionised litmus is red The ionised litmus is blue

LITMUS ADDING H + ions ADDING OH - IONS At some point during the movement of the position of equilibrium, the concentrations of the two colours will become equal. The colour you see will be a mixture of the two. 63

INDICATOR COLORS IN TITRATION Indicator Acid color Transition color Base color STRONG ACID STRONG BASE Litmus 2 3 4 5 6 7 8 9 10 11 12 ph Bromthymol blue

INDICATOR COLORS IN TITRATION Indicator Acid color Transition color Base color WEAK ACID STRONG BASE Phenolphthalein 2 3 4 5 6 7 8 9 10 11 12 ph Phenol red

INDICATOR COLORS IN TITRATION Indicator Acid color Transition color Base color STRONG ACID WEAK BASE Methyl orange 2 3 4 5 6 7 8 9 10 11 12 ph Bromphenol blue

THE ph RANGE OF INDICTORS Acid-base indicators take advantage of the rapid change in ph of the solution being titrated as the equivalence point is reached. Indicators dose not change colour sharply at one particular ph, they change over a narrow range of ph. Indicators ph Litmus 5-8 Methyl Orange 3.1-4.4 phenophthalein 8.3-10.0

EQUIVALENCE POINT END POINT Equivalence point (theoretical completion of the titration) : moles of titrant = moles of titrand. (stoichimoetric point or chemical equivalence) Endpoint is just the colour change from the indicator (physical). During a titration, the ph changes sharply close to the stoichiometric point and the indicator used MUST CHANGE COLOUR over the same range. 68

CHOOSING INDICATOR FOR TITRATION A suitable indicator has a color change that lies within the ph equivalence point. The way you normally carry out a titration involves adding the acid to the alkali. 69

TITRATION CURVE FOR STRONG BASE WITH STRONG ACID During a ph titration, the ph changes in a characteristic way. A ph curve is found of the ph of solution being titrated is plotted against the volume of solution added. Indicators ph Methyl orange 3.1-4.4 phenophthalein 8.3-10.0 Addition of the alkali to the acid The way you normally carry out a titration involves adding the acid to the alkali.

TITRATION CURVE FOR WEAK BASE V STRONG ACID This time we are going to use hydrochloric acid as the strong acid and ammonia solution as the weak base. Indicators ph Methyl Orange 3.1-4.4 phenophthalein 8.3-10.0

TITRATION OF WEAK BASE WITH STRONG ACID Indicator Methyl Orange Phenolphalein ph 3.1-4.4 8.3-10.0 M.O is useful phph is useless

TITRATION CURVES FOR STRONG BASE V WEAK ACID Indicators ph Methyl Orange 3.1-4.4 phenophthalein 8.3-10.0 Ph.ph is useful M.O is useless

TITRATION CURVES FOR WEAK BASE WITH WEAK ACID Indicators ph Methyl Orange 3.1-4.4 phenophthalein 8.3-10.0 Both php and MO are useless

SUMMARY